joj1349 Oil Deposits

1349
: Oil Deposits

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	533	260	Standard

The GeoSurvComp geologic survey company is responsible for detecting
underground oil deposits. GeoSurvComp works with one large rectangular
region of land at a time, and creates
a grid that divides the land into numerous square plots. It then analyzes
each plot separately,
using sensing equipment to determine whether or not the plot contains oil.

A plot containing
oil is called a pocket. If two pockets are adjacent, then they are part of
the same oil deposit. Oil
deposits can be quite large and may contain numerous pockets. Your job is to
determine how many different oil deposits are contained in a grid.

Input

The input file contains one or more grids. Each grid begins with a line
containing m
and n
, the
number of rows and columns in the grid, separated by a single space. If
m
= 0 it signals the end
of the input; otherwise



and



.
Following
this are m
lines of n
characters
each (not counting the end-of-line characters). Each character corresponds to
one plot, and is
either `*
', representing the absence of oil, or `@
', representing an oil pocket.

Output

For each grid, output the number of distinct oil deposits. Two different
pockets are part of the
same oil deposit if they are adjacent horizontally, vertically, or diagonally.
An oil deposit will not contain more than 100 pockets.

Sample Input

1 1

*

3 5

*@*@*

**@**

*@*@*

1 8

@@****@*

5 5

****@

*@@*@

*@**@

@@@*@

@@**@

0 0

Sample Output

0

1

2

2

基本的深搜，回溯法找联通块

#include<iostream>

#include<cstring>

using namespace std;

char map[105][105];

int vis[105][105];

int m,n,count1;

void dfs(int x,int y)

{

if(x>=1&&x<=m&&y>=1&&y<=n)

{

if(map[x][y]=='*'||vis[x][y]) return;

vis[x][y]=1;

dfs(x-1,y-1); dfs(x-1,y);dfs(x-1,y+1);

dfs(x,y-1);             dfs(x,y+1);

dfs(x+1,y-1); dfs(x+1,y);dfs(x+1,y+1);

}

else return;

}

int main()

{

while(cin>>m>>n)

{

if(m==0&&n==0) break;

count1=0;

memset(vis,0,sizeof(vis));

for(int i=1;i<=m;i++)

for(int j=1;j<=n;j++)

cin>>map[i][j];

for(int i=1;i<=m;i++)

for(int j=1;j<=n;j++)

if(!vis[i][j]&& map[i][j]=='@') {count1++;dfs(i,j);}

cout<<count1<<endl;

}

return 0;

}